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Let $G=\mathrm{Aut}(\mathbf C)$, the group of field automorphisms of the complex numbers. It is a very large group (see this MSE question and the nice answer by Andres). For instance, there even exist automorphisms mapping $\pi$ to $e$.

In Galois theory, one puts the profinite (Krull) topology on Galois groups of infinite extensions. This is very natural for a lot of reasons.

Does there exist a natural topology on $G$? It is not profinite, so the profinite topology is out of the question. However, after picking an embedding $\overline{\mathbf Q} \hookrightarrow \mathbf C$, the restriction map $\mathrm{Aut}(\mathbf C) \to \mathrm{Gal}(\overline{\mathbf Q}/\mathbf Q)$ is a map from $G$ to a profinite group. Is there a natural, intrinsically defined topology on $G$ for which this map is continuous? Of course, there are silly topologies which would do this (eg. the discrete topology, or the topology induced by the map), but I'm wondering whether there is one which is not silly.

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Bruno Joyal
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  • The fact that the Galois group of an algebraic extension is profinite is, in some sense, incidental. There are other ways of defining the Krull topology so that it makes sense more generally. – Zhen Lin May 03 '14 at 09:07
  • @ZhenLin I'm curious! If you'd like to elaborate in an answer, I'll gladly accept it. – Bruno Joyal May 07 '14 at 20:34

1 Answers1

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There is a general notion of Krull topology in model theory that can be applied to automorphism groups of any one-sorted structure, and in the special case of a algebraic field extension, this is the usual topology on the Galois group. The following facts are standard:

  • The automorphism group becomes a topological group under this topology, and its action on the model is continuous.
  • The topology is totally disconnected and Hausdorff.
  • The automorphism group is profinite if and only if every element of the model has a finite orbit under the action.

Now, given a field $K$ and an extension $L$, the Krull topology on $\mathrm{Aut}(L \mid K)$ is generated by the basic open subsets $\{ g \in \mathrm{Aut}(L \mid K) : g (\vec{a}) = \vec{a}' \}$, where $\vec{a}$ and $\vec{a}'$ are finite lists of elements of $L$ (of the same length). It is straightforward to see that this topology is natural in the sense of being compatible with restriction to Galois subextensions.

Zhen Lin
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